Give a concrete sequence of rationals which converges to an irrational number and vice versa. Give a concrete sequence of rationals which converges to an irrational number and vice versa....
My work
I could give a sequence of irrationals which converges to a rational number...
Let $r\in \mathbb Q,$ $$a_n=\frac {\sqrt 2} n+r$$
But I couldn't give a sequence of rationals which converges to an irrational.. Help me to work out.....
 A: $a_n = \left( 1 +\dfrac{1}{n} \right)^n $ converges to $e$
A: Take any irrational number, and look at its decimal expansion, like $$\sqrt{2}= 1.41421356237310\ldots$$
A sequence that converges to it is
$$1, 1.4, 1.41, 1.414, 1.4142, 1.41421, \ldots$$
Basically, just add a digit each time. This is clearly a sequence of rationals because every decimal number with a finite decimal expansion is rational, and it clearly converges to the irrational number in question (if it didn't, then decimal expansions would not be well-defined, but this can also be proved directly using geometric series). 
This seems like a trivial example, but understanding it is really fundamental to understanding how decimal expansions work, and what infinite decimal expansions really are.
A: There have been a lot of elegant and precise examples in response to this question, but I want to show that actually, you don't need elegance or precision: there are loads of possible answers to this kind of question.
Start with your candidate number, say $\sqrt 2$. Choose some range about it, say $\sqrt{2} \pm 1$. Now, that's a reasonable chunk of the number line, so you must be able to find a rational number in it, because rational numbers are packed in everywhere. $3/2$ will do, that's the first element of your sequence. Now choose a smaller range, say $\sqrt{2} \pm 1/2$. It turns out $3/2$ fits in this one too, so let's pick that one again. Now choose $7638$. That's nowhere near where you want to be, but it actually doesn't matter – if you only make finitely many "mistakes" like that, it doesn't make the slightest bit of difference to where you converge. 
Okay, now let's pick some rational in the range $1$ to $1.5$, because that's a smaller range than before that still contains $\sqrt{2}$. Let's pick $1.23523$ (notice this is actually further away than $3/2$ was – this doesn't matter). Narrow the range to $1.25$ to $1.45$ (you can check this range is still ok by squaring both ends and checking the lower one falls below $2$ and the upper one above it). Pick another rational. Narrow again, and pick again. Keep going forever.
As long as the width of the ranges you allow eventually approach zero, the rational sequence you pick will eventually converge to $\sqrt 2$.

Here's another approach: pick a sequence that converges to $0$. Add $\sqrt{2}$ to every element. Now your sequence converges to $\sqrt{2}$, but it might not have rational elements. However, you can fix it: as before, rationals are packed in everywhere, so you ought to be able to move each element of your sequence just a tiny amount – say, less than its distance to $\sqrt{2}$, so it ends up less than twice as far away – and hit a rational number nearby. Now all your sequence members are rational, and you're done.
A: How about $a_n$ = the decimal expansion of $\sqrt{2}$ up to the $n$-th place
A formal definition could be $$a_n = \lfloor 10^n\sqrt{2} \rfloor 10^{-n} $$
A: question 1: $3, 3.1, 3.14, 3.141, ...$
question 2: $3.14159...,  0.14159..., 0.04159...., 0.00159, ...$
A: $4*(\frac11-\frac13+\frac15-\frac17\dots)$ is known to converge to $\pi$ (Ramanujan).
A: Take ratios of consecutive Fibonacci numbers: $\frac11,\frac21,\frac32,\frac53,\frac85,\dots$.  It is well known that this converges to the golden ratio $\frac{1+\sqrt{5}}{2}$, which is irrational.
A: Consider the sequence
$$0.1, 0.101, 0.101001, 0.1010010001, 0.101001000100001, 0.101001000100001000001,\dots.$$
This converges to $\alpha=0.101001000100001000001\cdots$. The decimal expansion of $\alpha$ is not ultimately periodic, so $\alpha$ is irrational.
A: Any Taylor series that converges to an irrational would work. E.g.
$$\frac{1}{0!}+\frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots = e$$
Pick any irrational number, pick a function that calculates it given a rational input, then the taylor series of that function around that input will fulfill the requirements.
A: Let $x$ be the irrational number you are trying to obtain as the limit of a sequence of rationals. Let's assume that you can calculate the decimal expression of $x$ to the $n$th term.
Then, you can define a sequence of rationals where the $n$th term is the number $x$ with $n$ decimal digits. A number with a finite number of decimal digits is a rational number (you can multiply and divide by ten to the $n$, being $n$ the number of decimals).
It's obvious that this sequence converges to your irrational number, and the definition can easily be formalized.
With this method you can obtain ANY irrational number as long as you can calculate the $n$th decimal term with an algorithm.
A: Let $x_n = {1 \over n}\sqrt{2}$. $x_n \rightarrow 0.$ $x_n$ is irrational for each $n \in \Bbb N$ but $0$ is rational.
Consider the decimal expansion of $\sqrt 2 = 1.414213562$; let $q_n$ but that decimal expansion truncated to $n$ decimal places.
Alternatively, you (may) know that $\sin(x) = \sum_{r=0}^\infty {(-1)^r x^{2r+1} \over (2r+1)!}$; let $s_n = \sum_{r=0}^n {(-1)^r ({1 \over 3}\pi)^{2r+1} \over (2r+1)!}$. $s_n \rightarrow \sin({1 \over 3}\pi) = {1 \over 2}\sqrt{2}$.
Hopefully these are slightly different to some of the previous answers! :)
A: Let $a_1=1$ and recursively $a_{n+1}=\frac{a_n}2+\frac{1}{a_n}$ and show that this converges to $\sqrt 2$.
A: Set $a_0=1$ and
$$
a_{n+1}=\frac{3a_n+4}{2a_n+3}
$$
and prove the sequence is increasing and bounded; then prove that it converges to $\sqrt{2}$.
A: $$
1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 \ddots}}}}
$$
The simple continued fraction expansion of a rational number must terminate because you can't keep getting smaller and smaller positive integers, thus, for example:
\begin{align}
\frac{67}{30} & = 2 + \frac 7 {30} & & (\text{7 is smaller than 30}) \\[10pt]
& = 2 + \frac 1 {\left(\frac{30} 7 \right)} = 2 + \cfrac 1 {4 + \cfrac 2 7} & & (\text{2 is smaller than 7}) \\[10pt]
& = 2 + \cfrac 1 {4 + \cfrac 1 {\left(\frac 7 2 \right)}} = 2 + \cfrac 1 {4 + \cfrac 1 {3 + \cfrac{1}{2}}} & & (\text{1 is smaller than 2})
\end{align}
Hence the thing that an infinite simple continued fraction converges to must be irrational.
A: Take any positive number $\alpha \notin \mathbb Q$. Then the rational sequence $a_n = \dfrac{\lfloor n\alpha \rfloor}{n}$ converges to $\alpha$.
(This is also true for $\alpha \in \mathbb Q$, but that case is not relevant to your question.)
A: You can get this result using Fourier series:
$$
\pi = 4\sum\limits_{n=1}^\infty \frac{(-1)^{n+1}}{2n - 1}
$$
A: One way to go about this is to use newton's method combined with polynomials.
For instance, we know that the roots of $$f(x) = x^2 - 2$$ are $\pm \sqrt{2}$. We seed our sequence with $2$ as a first guess. The next guess comes from newton's method, this is where we compute the linearization of $f(x)$ at $2$ and find where the linearization is zero. This becomes our new guess: $$L(x) = f'(2)(x-2)+f(2) = 4(x-2) + 2$$ and we see that $L(x) = 0$ when $x = 1.5$. Thus the next term in the sequence is $1.5$.
Given a point $x_n$ we can find the next point $x_{n+1}$ via: $$L(x_{n+1}) = 0 = f'(x_n)(x_{n+1}-x_n)+f(x_n).$$ Solving for $x_{n+1}$, we have $$x_{n+1} = \frac{-x_n^2+2}{2x_n} + x_n = \frac{x_n^2+2}{2x_n} = \frac{x_n}{2} + \frac{1}{x_n}.$$
It isn't hard to prove convergence of the series, given that you choose good initial conditions. (Say $x_0 \ge 2$) Notice that this is the same sequence proposed by @HagenvonEitzen.
A: Quite a few previous answers have give sequences of rationals converging to an irrational.  Here's a  sequence of irrational numbers converging to a positive rational $r$:
Let $a_0=0$ and $s_i=\Sigma_0^i a_i$ and $a_{i+1}= (r-s_i)/\pi$.
(A flaw in this method: some values among the $a_i$ might by chance be rational; the sum of two irrationals need not be irrational.  To address this flaw, I need to add a rational fraction of an irrational to  $a_i$, rather than adding an irrational fraction of an irrational to  $a_i$.  This can be done by a technique similar to that in the next part of this answer.)
A sequence of rationals converging to a positive irrational $x$ can be created in a related way:
Let $a_0=0$ and $a_{i+1} = a_i + 1/k$, with $k$ the largest integer such that $a_{i+1} < x$.
A: Consider $x_n=\frac{\sqrt{a}\left[(\sqrt{a}+1)^{2n}+(\sqrt{a}-1)^{2n}\right]}{(\sqrt{a}+1)^{2n}-(\sqrt{a}-1)^{2n}}$. 
It is rather straightforward to see that this sequence converges to $\sqrt{a}$. The 'surprise' or key point is that actually $x_n$ is rational when $a$ is rational. This follows immediately upon using the binomial expansion.
As an example consider $a=2$. One then finds $x_n=(\frac{3}{2},\frac{17}{12},\frac{99}{70},\ldots)=(1.5, 1.417\ldots,1.414\ldots,\ldots)$, already pretty good approximations of $\sqrt{2}=1.4142\ldots$
Note: This example is essentially a reworking of the Fibonnaci to golden ratio example.
A: Let $c$ be an irrational number. Now we get a rational between $c-1/n$ and $c+1/n$ for all natural numbers $n$. Then, we consider the sequence $x_n$, of the  rational numbers between $c-1/n$ and $c+1/n$. Then, by squeeze theorem, we get $\lim (x_n)=c$.
A: Let $(u_{n})_{n\geq1}$ be the sequence defined by$$u_{n}=\frac{1}{n}+\dots+\frac{1}{2n}.$$
Then ${\displaystyle \lim_{n\to+\infty}u_{n}=\ln2}$.
We can prove this using the inequalities
$$\frac{1}{x+1}<\ln\left(x+1\right)-\ln\left(x\right)<\frac{1}{x}.$$
